Thanks default, I will look at the links you offerd. You mentioned
using MOSFETs, I may not be able to use these because I need to drive
two different base(gate) with less than 25mA from a PIC Output signal
that also goes through a voltage drop (due to a diode) to about 3.8v.
So the MOSFETS, without a transistor amplifier, are out of the
question.
At 25mA, the PIC outputs probably won't be 4.5V (assuming a 5V supply,
about which I'm also not sure.) My guess is that they will be about
25mA*70ohms or 1.75V less than your supply voltage before they meet
your diode. (Actual impedance will vary some and you need to be aware
that you cannot drive all pins at the same 25mA, as there is another
spec that provides an absolute maximum [don't get too close to it]
that limits the total port output.) Or were you thinking that it was
a diode there in the PIC to explain the 3.8V output?
Anyway, I think we understand the following:
Drive signal: PIC output
Number of circuits: 2
Mode of operation: switch (unless I assume far too much here)
Load current: 220mA
Load impendance: unknown, but may have significant L, if relay
Load voltage rail: unknown
PIC Voltage rail: ??, but either 4.5V (calculated from your
assumed diode drop and 3.8V mentioned) or
else 5V (blindly assumed as "normal")
Could you expand the above to provide more precise information? A
highly inductive or highly capacitive load may require other caveats
as compared to a purely resistive load. Also, are you planning on
connecting the load between ground and the 'switch circuit' or between
some power rail and the 'switch circuit'?
Assuming the above doesn't stray too far from your situation, one note
right now is that you should want to keep your PIC drive current as
low as is reasonable, while being able to reliably drive the switch
circuit. I'd target _much_ less than 25mA as reasonable, here.
Something on the order of just a few mA. I don't like the idea of
just seeing the maximum spec of 25mA and merrily planning on using
that as an option.
Jon